An insulated 0.2 m³ tank contains helium at 1200 kPa and 47^oC. A valve is now opened, allowing some helium to escape. The valve is closed when one-half of the initial mass has escaped. The temperature of the gas is $(\sqrt[3]{4}=1.6)$.

Initial mass of air, $\mathrm{m}=\frac{{\mathrm{P}}_1\mathrm{V}}{{\mathrm{RT}}_1}$

Final mass of air

   $\begin{array}{l}=\frac{\mathrm{m}}{2}=\frac{{\mathrm{P}}_2\mathrm{V}}{{\mathrm{RT}}_2}\\ \frac{{\mathrm{P}}_1\mathrm{V}}{2{\mathrm{RT}}_1}=\frac{{\mathrm{P}}_2\mathrm{V}}{{\mathrm{RT}}_2}\end{array}$

     $\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)=2\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)$

As the tank is insulated, the process is adiabatic with $\mathrm{\gamma }=\frac{5}{3}$

    $\begin{array}{c}{\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)}^{\frac{\mathrm{\gamma }-1}{\mathrm{\gamma }}}\Rightarrow {\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\frac{5}{2}}=\left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)\\ \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}=2{\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\frac{5}{2}}\Rightarrow {\mathrm{T}}_2=\frac{{\mathrm{T}}_1}{2^{\frac{2}{3}}}\\ =\frac{320}{2^{\frac{2}{3}}}=200\mathrm{K}=-{73}^{\circ }\mathrm{C}[\text{ Given }\sqrt[3]{4}=1.6]\end{array}$